Question

Consider following equilibriums and their equilibrium constants of lysine.
$H_3{N}^{+}-\left(C{H}_2{)}_{4}CH\right(N{H}_3^{+})COOH\stackrel{pK=2.2}{\rightleftharpoons }{H}_3{N}^{+}-(C{H}_2{)}_{4}CH\left(N{H}_3^{+}\right)CO{O}^{-}+H^{+}$
$H_3{N}^{+}-\left(C{H}_2{)}_{4}CH\right(N{H}_2)CO{O}^{-}+H_{2}O\stackrel{pK=5}{\rightleftharpoons }{H}_3{N}^{+}-(C{H}_2{)}_{4}CH\left(N{H}_3^{+}\right)CO{O}^{-}+O{H}^{-}$

$H_{2}N-\left(C{H}_2{)}_{4}CH\right(N{H}_2)CO{O}^{-}+H_{2}O\stackrel{pK=3.4}{\rightleftharpoons }{H}_3{N}^{+}-(C{H}_2{)}_{4}CH\left(N{H}_2\right)CO{O}^{-}+O{H}^{-}$

Answer 1

Since, $p{K}_a+p{K}_b=14$
Hence
$H_3{N}^{+}-\left(C{H}_2{)}_{4}CH\right(N{H}_3^{+})COOH\stackrel{p{K}_1=2.2}{\rightleftharpoons }{H}_3{N}^{+}-(C{H}_2{)}_{4}CH\left(N{H}_3^{+}\right)CO{O}^{-}+H^{+}$

$H_3{N}^{+}-\left(C{H}_2{)}_{4}CH\right(N{H}_3^{+})CO{O}^{-}\stackrel{p{K}_2=9}{\rightleftharpoons }{H}_3{N}^{+}-(C{H}_2{)}_{4}CH\left(N{H}_2\right)CO{O}^{-}+H^{+}$

$H_3{N}^{+}-\left(C{H}_2{)}_{4}CH\right(N{H}_2)CO{O}^{-}\stackrel{p{K}_3=10.6}{\rightleftharpoons }{H}_{2}N-(C{H}_2{)}_{4}CH\left(N{H}_2\right)CO{O}^{-}+H^{+}$

So, at isoelectric point:

$pI=\frac{p{K}_2+p{K}_3}{2}=\frac{9+10.6}{2}=9.8$